Find the equation of the line that is perpendicular to the graph of 3x + 2y = 12 and contains the point (2, -3), then write the equation of the line in slope-intercept form.
The Correct Answer and Explanation is:
To find the equation of the line that is perpendicular to the graph of 3x+2y=123x + 2y = 123x+2y=12 and contains the point (2,−3)(2, -3)(2,−3), we follow these steps:
Step 1: Find the slope of the given line
First, we need to rewrite the equation 3x+2y=123x + 2y = 123x+2y=12 in slope-intercept form, which is y=mx+by = mx + by=mx+b, where mmm is the slope of the line.
Start by isolating yyy in the equation: 3x+2y=123x + 2y = 123x+2y=12
Subtract 3x3x3x from both sides: 2y=−3x+122y = -3x + 122y=−3x+12
Now, divide both sides by 2: y=−32x+6y = -\frac{3}{2}x + 6y=−23x+6
The slope of the given line is m=−32m = -\frac{3}{2}m=−23.
Step 2: Find the slope of the perpendicular line
The slope of a line that is perpendicular to another is the negative reciprocal of the original slope. The negative reciprocal of −32-\frac{3}{2}−23 is 23\frac{2}{3}32.
So, the slope of the perpendicular line is 23\frac{2}{3}32.
Step 3: Use the point-slope form to write the equation
We know the slope of the perpendicular line is 23\frac{2}{3}32 and that it passes through the point (2,−3)(2, -3)(2,−3). We can use the point-slope form of the equation of a line: y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1=m(x−x1)
where m=23m = \frac{2}{3}m=32 and (x1,y1)=(2,−3)(x_1, y_1) = (2, -3)(x1,y1)=(2,−3).
Substitute the values: y−(−3)=23(x−2)y – (-3) = \frac{2}{3}(x – 2)y−(−3)=32(x−2)
Simplify: y+3=23(x−2)y + 3 = \frac{2}{3}(x – 2)y+3=32(x−2)
Now, expand the right side: y+3=23x−43y + 3 = \frac{2}{3}x – \frac{4}{3}y+3=32x−34
Step 4: Isolate yyy to get the slope-intercept form
Subtract 3 from both sides: y=23x−43−3y = \frac{2}{3}x – \frac{4}{3} – 3y=32x−34−3
Rewrite 3 as 93\frac{9}{3}39: y=23x−43−93y = \frac{2}{3}x – \frac{4}{3} – \frac{9}{3}y=32x−34−39
Combine the constants: y=23x−133y = \frac{2}{3}x – \frac{13}{3}y=32x−313
Final Answer:
The equation of the line that is perpendicular to 3x+2y=123x + 2y = 123x+2y=12 and contains the point (2,−3)(2, -3)(2,−3) is: y=23x−133y = \frac{2}{3}x – \frac{13}{3}y=32x−313
